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tags: data_structure_python
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# Perform String Shifts <img src="https://img.shields.io/badge/-easy-brightgreen">

You are given a string s containing lowercase English letters, and a matrix shift, where shift[i] = [direction, amount]:

- direction can be 0 (for left shift) or 1 (for right shift).
- amount is the amount by which string s is to be shifted.
- A left shift by 1 means remove the first character of s and append it to the end.
- Similarly, a right shift by 1 means remove the last character of s and add it to the beginning.

Return the final string after all operations.

<ins>**Example 1:**</ins>

Input: s = "abc", shift = [[0,1],[1,2]]
Output: "cab"
Explanation: 
[0,1] means shift to left by 1. "abc" -> "bca"
[1,2] means shift to right by 2. "bca" -> "cab"

<ins>**Example 2:**</ins>

Input: s = "abcdefg", shift = [[1,1],[1,1],[0,2],[1,3]]
Output: "efgabcd"
Explanation:  
[1,1] means shift to right by 1. "abcdefg" -> "gabcdef"
[1,1] means shift to right by 1. "gabcdef" -> "fgabcde"
[0,2] means shift to left by 2. "fgabcde" -> "abcdefg"
[1,3] means shift to right by 3. "abcdefg" -> "efgabcd"

**Constraints:**

- 1 <= s.length <= 100
- s only contains lower case English letters.
- 1 <= shift.length <= 100
- shift[i].length == 2
- 0 <= shift[i][0] <= 1
- 0 <= shift[i][1] <= 100

# Solution

### Solution 1: 1st approach

class Solution:
    def stringShift(self, s: str, shift: List[List[int]]) -> str:
        totalShift = 0
        for direction, amount in shift:
            if direction == 0:
                totalShift += amount
            else:
                totalShift -= amount
        
        return s[totalShift % len(s):] + s[:totalShift % len(s)]

### Solution 2: 2nd approach

class Solution:
    def stringShift(self, s: str, shift: List[List[int]]) -> str:
        r = sum((1-d*2)*a for [d,a] in shift) %len(s)
        return s[r:]+s[:r]